Suppose I have two square matrices of the same dimension, say $A_1, A_2$ over the complex numbers. The Kronecker sum is defined by $A_1 \oplus A_2 = A_1 \otimes I + I \otimes A_2$, and moreover $A_1 \otimes I$ and $A_2 \otimes I$ commute. Therefore $e^{A_1\oplus A_2} = e^{A_1 \otimes I + I \otimes A_2} = e^{A_1 \otimes I}e^{I \otimes A_2}$. Now in a paper I am reading this last equality is then written as $(e^{A_1} \otimes I) (I \otimes e^{A_2})$, which I am assuming means the identities $e^{A \otimes I} = e^{A} \otimes I$ and $e^{I \otimes A} = I \otimes e^A$ are valid.
Considering the first identity: $A \otimes I = \begin{bmatrix} a_{1,1}I & a_{1,2}I &\cdots &a_{1,n}I \\ a_{2,1}I & a_{2,2}I &\cdots & a_{2,n}I \\ \cdots \\ a_{n,1}I & a_{n,2}I &\cdots &a_{n,n}I \end{bmatrix}$, and $e^{A \otimes I} = \sum_{k=0}^\infty \frac {1} {k!} (A\otimes I)^k$, and I am not sure how to show that this identity holds (and hopefully this is not just a typo in the paper).
You have $$(A\otimes B)^n = A^n \otimes B^n$$ so $$ \exp(A\otimes I) = \sum_{n=0}^\infty \frac{1}{n!}(A\otimes I)^n = \sum_{n=0}^\infty \frac{1}{n!} A^n \otimes I = \exp(A) \otimes I$$